Differentiating Exponential and Logarithmic Functions

Differentiaing Exponentials

The exponential function \(e^x\), when differentiated, gives itself. This is shown as the rule below:

\[ \textrm{If }y=e^x\textrm{, then }\frac{dy}{dx} = e^x \]

We can also differentiate when the \(x\) has a coefficient, for example \(y=e^{4x}\), and also when the exponential is multiplied by a constant, for example \(y=3e^x\). We combined these into the rule below:

\[ \textrm{If }y=ae^{bx}\textrm{, then }\frac{dy}{dx} = a\times b\times e^{bx} \]

Example

Differentiate the following with respect to \(x\):

1. \(y=e^2x\)

2. \(y=8\exp{\left(-\frac{x}{3}\right)}\)

Solution:

1. Using the rule:

\[ \textrm{If }y=ae^{bx}\textrm{, then }\frac{dy}{dx} = a\times b\times e^{bx} \]

note \(y=e^2x\), so \(a=1\) and \(b=2\), hence \(\frac{dy}{dx}=2e^{2x}\).

2. Using the rule:

\[ \textrm{If }y=ae^{bx}\textrm{, then }\frac{dy}{dx} = a\times b\times e^{bx} \]

note \(y=8\exp{\left(-\frac{x}{3}\right)}\), so \(a=8\) and \(b=-\frac{1}{3}\), hence

\[ \frac{dy}{dx}=8\times -\frac{1}{3}\times \exp{\left(-\frac{x}{3}\right)} = -\frac{8}{3}\exp{\left(-\frac{x}{3}\right)}. \]

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We can check these results with sympy.

from sympy import symbols, diff, exp

x = symbols('x') 

diff(exp(2 * x))

\[\displaystyle 2 e^{2 x}\]

diff(8 * exp(-x / 3))

\[\displaystyle - \frac{8 e^{- \frac{x}{3}}}{3}\]

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Differentiating Logarithms

We differentiate logarithmic functions using the rule shown below:

\[ \textrm{If }y=\ln{(x)}\textrm{, then }\frac{dy}{dx} = \frac{1}{x} \]

We can differentiate when the \(x\) has a coefficient, for example \(y=\ln{(4x)}\), and also when the logarithmic function has a coefficient, for example \(y=3\ln{(x)}\). We combine these into the rules below:

Note

In this rule, the \(b\) disappears and it forms a nice exercise in the practice of using the laws of logarithms, hence we have included the algebra in the next example. We do have to differentiate a sum however, so it you are unsure on that step, check ou the next section for more help.

\[ \textrm{If }y=a\ln{(bx)}\textrm{, then }\frac{dy}{dx} = \frac{a}{x} \]

Example

For \(y=a\ln{(bx)}\), find \(\frac{dy}{dx}\).

Solution: Using laws of logarithms, we can write \(y=a\ln{(bx)}\) as:

\[ y = a\left[\ln{(x)} + \ln{(b)}\right] \]

Now differentiating this as a sum gives:

\[ \frac{dy}{dx} = \frac{d}{dx}\left\{a\left[\ln{(x)} + \ln{(b)}\right]\right\} = a\frac{d}{dx}\left[\ln{(x)}\right] + a\frac{d}{dx}\left[\ln{(b)}\right] \]

Now, as \(\frac{d}{dx}\left[\ln{(x)}\right] = \frac{1}{x}\) and \(\frac{d}{dx}\left[\ln{(b)}\right] = 0\), as \(\ln{(b)}\) is a constant, so we have:

\[ \frac{dy}{dx} = \frac{a}{x} \]

Example

Differentiate:

1. \(y = \ln{(3x)}\)

2. \(y = -\frac{4}{5}\ln{\left(\frac{x}{2}\right)}\)

Solution:

1. We use the rule:

\[ \textrm{If }y=a\ln{(bx)}\textrm{, then }\frac{dy}{dx} = \frac{a}{x} \]

Note that we have \(y=\ln{(3x)}\), so \(a=1\) and \(b=3\). Hence \(\frac{dy}{dx}=\frac{1}{x}\).

2. We use the rule:

\[ \textrm{If }y=a\ln{(bx)}\textrm{, then }\frac{dy}{dx} = \frac{a}{x} \]

Note that we have \(y=-\frac{4}{5}\ln{\left(\frac{x}{2}\right)}\), so \(a=-\frac{4}{5}\) and \(b=\frac{1}{2}\). Hence \(\frac{dy}{dx} = -\frac{4}{5}\times \frac{1}{x} = -\frac{4}{5x}\).

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Again, sympy can be used to show this also.

Python Note

Similar to numpy, sympy uses log for a natural logarithm and log10 for a base-10 logarithm.

from sympy import log

diff(log(3 * x))

\[\displaystyle \frac{1}{x}\]

diff(-4 / 5 * log(x / 2))

\[\displaystyle - \frac{0.8}{x}\]